Inequality on sines and factorial

Algebra Level 3

Find the number of solutions to the inequality E = 2 1 sin ⁡ 2 θ 2 ⋅ 3 1 sin ⁡ 2 θ 3 ⋅ 4 1 sin ⁡ 2 θ 4 ⋯ n 1 sin ⁡ 2 θ n ≤ n ! \large E= 2^{ \frac{1}{\sin^{2} \theta_{2}}}\cdot 3^{ \frac{1}{\sin^{2} \theta_{3}}} \cdot 4^{ \frac{1}{\sin^{2} \theta_{4}}} \cdots n^{ \frac{1}{\sin^{2} \theta_{n}}} \le n! where θ i ∈ ( − π , 2 π ) \theta_{i} \in (-\pi, 2\pi) for i = 2 , 3 , … , n i = 2,3,\ldots, n .

0 2 n − 1 2^{n-1} 2 n 2^{n} None of the above

Baibhab Chakraborty by Baibhab Chakraborty , 16, India

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1 solution

sin ⁡ θ ϵ [ − 1 , 1 ] \sin \theta \epsilon [-1,1]

sin ⁡ 2 θ ϵ [ 0 , 1 ] \sin^{2} \theta \epsilon [0,1]

1 sin ⁡ 2 θ ϵ [ 1 , α ] \frac{1}{\sin^{2} \theta} \epsilon [1, \alpha]

Since minimum possible value of 1 sin ⁡ 2 θ \frac{1}{\sin^{2} \theta} is 1 , and 2.3.4.....n=n!, thus the solution to this can be only sin ⁡ 2 θ i = 1 \sin^{2} \theta_{i} =1

sin ⁡ θ i = ± 1 \sin \theta_{i} = \pm 1

When θ i ϵ ( − π , 2 π ) \theta_{i} \epsilon (-\pi, 2\pi) there can be three values of θ \theta corresponding to s i n θ = ± 1 sin \theta =\pm 1 .

Thus the number of solutions = 3 n − 1 3^{n-1} .

Please post better solutions to this problem, I am a novice.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Please change

for i=1,2,3,4...n

To

for i=2,3,4...n

Because with the first it is clear there are infinite solutions, when the problem intends different

Jason Gomez - 2 months, 2 weeks ago

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